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author | David A. Madore <david+git@madore.org> | 2016-02-16 15:46:21 +0100 |
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committer | David A. Madore <david+git@madore.org> | 2016-02-16 15:46:21 +0100 |
commit | 362ce5d5dc7aa081696c0707d52111947a69c31a (patch) | |
tree | 5378f523c4329cd88d59d88953522287227e71cd | |
parent | 19f2c053fe5485d4246568ce4a7b93ff324eabf6 (diff) | |
download | mitro206-362ce5d5dc7aa081696c0707d52111947a69c31a.tar.gz mitro206-362ce5d5dc7aa081696c0707d52111947a69c31a.tar.bz2 mitro206-362ce5d5dc7aa081696c0707d52111947a69c31a.zip |
Illustration for Hercules-versus-Hydra game.
-rw-r--r-- | notes-mitro206.tex | 63 |
1 files changed, 62 insertions, 1 deletions
diff --git a/notes-mitro206.tex b/notes-mitro206.tex index f50a49d..11fd8c2 100644 --- a/notes-mitro206.tex +++ b/notes-mitro206.tex @@ -18,7 +18,7 @@ \usepackage{graphics} \usepackage[usenames,dvipsnames]{xcolor} \usepackage{tikz} -\usetikzlibrary{matrix} +\usetikzlibrary{matrix,calc} % \theoremstyle{definition} \newtheorem{comcnt}{Tout}[subsection] @@ -647,6 +647,67 @@ souhaite. Hercule gagne s'il réussit à décapiter le dernier nœud de l'hydre ; l'hydre gagnerait si elle réussissait à survivre indéfiniment. +\begin{center} +\begin{tikzpicture}[baseline=0] +\draw[very thin] (-1.5,0) -- (1.5,0); +\begin{scope}[every node/.style={circle,fill,inner sep=0.5mm}] +\node (P0) at (0,0) {}; +\node (P1) at (0,1) {}; +\node (P2) at (-1,2) {}; +\node (P3) at (0,2) {}; +\node (P4) at (1,2) {}; +\node (P5) at (0.5,3) {}; +\node (P6) at (1.5,3) {}; +\end{scope} +\begin{scope}[line width=1.5pt] +\draw (P0) -- (P1); +\draw (P1) -- (P2); +\draw (P1) -- (P3); +\draw (P1) -- (P4); +\draw (P4) -- (P5); +\draw (P4) -- (P6); +\end{scope} +\begin{scope}[line width=3pt,red] +\draw ($(P6) + (-0.2,-0.2)$) -- ($(P6) + (0.2,0.2)$); +\draw ($(P6) + (-0.2,0.2)$) -- ($(P6) + (0.2,-0.2)$); +\end{scope} +\node[anchor=west] at (P6) {$x$}; +\node[anchor=west] at (P4) {$y$}; +\node[anchor=west] at (P1) {$z$}; +\end{tikzpicture} +devient +\begin{tikzpicture}[baseline=0] +\draw[very thin] (-1.5,0) -- (1.5,0); +\begin{scope}[every node/.style={circle,fill,inner sep=0.5mm}] +\node (P0) at (0,0) {}; +\node (P1) at (0,1) {}; +\node (P2) at (-1,2) {}; +\node (P3) at (0,2) {}; +\node (P4) at (0.8,2) {}; +\node (P5) at (0.6,3) {}; +\node (P4b) at (1.2,2) {}; +\node (P5b) at (1.2,3) {}; +\node (P4c) at (1.7,2) {}; +\node (P5c) at (1.7,3) {}; +\node (P4d) at (2.2,2) {}; +\node (P5d) at (2.2,3) {}; +\end{scope} +\begin{scope}[line width=1.5pt] +\draw (P0) -- (P1); +\draw (P1) -- (P2); +\draw (P1) -- (P3); +\draw (P1) -- (P4); +\draw (P4) -- (P5); +\draw[blue] (P1) -- (P4b); +\draw[blue] (P4b) -- (P5b); +\draw[blue] (P1) -- (P4c); +\draw[blue] (P4c) -- (P5c); +\draw[blue] (P1) -- (P4d); +\draw[blue] (P4d) -- (P5d); +\end{scope} +\end{tikzpicture} +\end{center} + Ce jeu est particulier en ce que, mathématiquement, non seulement Hercule possède une stratégie gagnante, mais en fait Hercule gagne \emph{toujours}, quoi qu'il fasse et quoi que fasse l'hydre. |